Consider the following statement:
Let $f\in\mathcal{C}^1(\mathbb{R}\to\mathbb{R})$ and $\lim_{x\to\infty}f(x)$ exists, then $\lim_{x\to\infty}f'(x)=0.$
I'm not sure whether my proof is correct:
Assume $h:=\lim_{x\to\infty}f'(x)\neq0$, without loss of generality $h>0$. So there exists $x_0>0$ so that $\forall x\geq x_0: f'(x)>h/2.$ From the convergence of $f$ follows $$\forall \epsilon>0\exists N\in\mathbb{N}\forall n\geq N: |f(n)-f(n+1)|<\epsilon.$$ From Taylor follows that there is a sequence $x_n\in[n,n+1]$ with $f'(x_n)=f(n+1)-f(n)$. For $\epsilon=h/2$ we get for all $n$ big enough $$h/2>|f(n)-f(n+1)|=|f'(x_n)|>h/2,$$ a contradiction.
EDIT: We have to make the additional assumption, that $f'$ converges. So the statement should be:
Let $f\in\mathcal{C}^1(\mathbb{R}\to\mathbb{R})$ and $\lim_{x\to\infty}f(x)$, $\lim_{x\to\infty}f'(x)$ exist, then $\lim_{x\to\infty}f'(x)=0.$
Let $f:\mathbb{R}\to\mathbb{R}$, $f(0)=1$ and $f(x)=\sin(e^x-1)/x$ otherwise. Then $f$ is continuously differentiable and $\lim_{x\to\infty}f(x)=0$, but $\lim_{x\to\infty}f'(x)$ doesn't exist.